perl regular expression take out text enclosed in parentheses - regex

Question

how do I use Perl to get rid of text within parentheses? For example:
$str = "This is a (extra stuff) string."
to
$str = "This is a string."
I am current using this but it's not working:
$str =~ s/( ( [^)]+ ) )//;
Thanks!

Answer 1

You need to escape the parentheses, like:
s/\([^)]*\)//g
Update by popular demand:
To remove the space you can simply remove spaces before the parenthesis. This will work in most cases:
s/\s*\([^)]*\)//g
To handle nested parenthesis you can use a recursive pattern, like so:
s/\s*\((?:[^()]+|(?R))*\)//g
You can read about (?R) and the like in perlre.
The last expression will work for string like aaa (foo(b,a,2*(3+4)) b) (c (c) c) ddd (x)., giving aaa ddd..

Answer 2

The ( are special and must be escaped
s/\([^)]+\)//g

Answer 3

None of the solutions so far do that the OP asked.
The expression $str =~ s/\([^)]*\)//g;
Converts "This is a (extra stuff) string" to "This is a string", leaving two spaces between the "a" and "string".
Converts "This is a (doubly (nested)) string" to "This is a ) string".
Converts "This is a (doubly (no, (triply!) nested) expression) string" to "This is a nested) expression) string".
Similar problems exist with $str =~ s/[ ]?\(.*?\)[ ]?//g; And why use those square brackets? Aren't regular expressions hairy enough without unneeded stuff?
We're going to need something a bit hairier to so we can eat multiply-nested parenthetical remarks and properly deal with keeping spacing where needed but discarding it otherwise. This does the trick:
1 while $str =~ s/(\w?)(\s*)\([^()]*\)(\s*)(\w?)
/($1&&$4)?($1.($2?$2:$3).$4):($1?$1:$4)/ex;
Edit
Test results:
'This string is OK as is.' -> 'This string is OK as is.'
'This is a (extra stuff) string.' -> 'This is a string.'
'(Preliminary remark) string' -> 'string'
'String (with end remark)' -> 'String'
'A string (remark before punctuation)!' -> 'A string!'
'A (doubly (nested)) string' -> 'A string'
'A (doubly (no, (triply!) nested)) string' -> 'A string'
Edit2
The exg qualification results in incorrect handling of "This (delete) (delete) is a string". All that is needed is ex.

Answer 4

This line should do what you need:
$str =~ s/[ ]?\(.*?\)[ ]?//g;
Do note that it won't work with nested brackets (like (this)), since the regex would have to be a lot more complicated for that type of functionality.

Answer 5

I do converting special characters to hex for easy use in my regex's
/\x28([^\x29]+)\x29/

Answer 6

Hmm I had expected the "greedy" principle to apply, eating all the way to the close parenthesis even when nested. Perhaps a little brute force, using index and rindex functions, would be better.
But I still wonder, why doesn't
$str =~ s/[ ]?\(.*?\)[ ]?//g;
slurp it all the way to the last ')'?

Answer 7

A split version. I kind of like split for this, because it is non-invasive, preserving the original format, and also, regexes tend to become... complicated. Though you need regex to trim it, of course.
You'd still need to work out the spacing. It is not a simple thing to predict whether extra space will appear in the front or end, and removing all double spaces will not preserve original format. This solution removes a single space in front of opening parens, and nothing else. Works in most cases, assuming the input has correct punctuation to begin with.
use warnings;
use strict;
while (<DATA>) {
my #parts = split /\(/;
print de_paren(#parts);
}
sub de_paren {
my $return = shift;
my #parts = #_;
while (my $word = shift #parts) {
next unless $word =~ /\)/;
$word =~ s/^.*?\)// while ($word =~ /\)/);
$return =~ s/ $//;
$return .= $word;
}
return $return;
}
__DATA__
A (doubly (no, (triply!) nested)) string
This is a (extra stuff) string.
(Preliminary remark) string
String (with end remark) String (with end remark)
A string (remark before punctuation)!
A (doubly (nested)) string
Output is:
A string
This is a string.
string
String String
A string!
A string ->